Given the arithmetic sequence: u1 = 124, u2 = 117, u3 = 110, u4 = 103 u(k) is the first term in the sequence that is negativefind the value of k

An arithmetic sequence is a sequence where every two consecutive terms differ by a fixed quantity - the common difference.

In this case, you can see that the next term is always 7 less than the previous one: 117 is seven less than 124, 110 is seven less than 117, and so on.

So, it makes sense to expect that this sequence will become negative at some time, because you're constantly taking away 7.

So, we're wondering how many times we can subtract 7 from 124 until it becomes negative:

So, the first integer solution is 18. This means that we have to subtract seven 18 times to get the first negative term.

Now, observe that we start from u1, so to build u2 we subtract 7 once, to build u3 we subtract 7 twice, and so on: the number of times we subtract 7 is shifted with respect to the sequence index.

So, we subtract seven 18 times to build u19, which is the first negative term.

1,8,15,22,29,36,43,50

step-by-step explanation: